If the line voltage of the power supply is 380V and the load impedance is 110 Ω, the line current and phase current of the circuit can be calculated

If the line voltage of the power supply is 380V and the load impedance is 110 Ω, the line current and phase current of the circuit can be calculated

If the line voltage is 380V, the phase voltage is 660V. If the single-phase impedance is 110 ohm, the phase current is 18a, and the line voltage is 660V
If the total three-phase impedance is 110 ohm, the phase current is 11.4a
Current 54a, line current 31a

It is known that the three-phase power line voltage UL = 380V, triangle connection symmetrical load z = (12 + j16) Ω. Calculate the load phase current and PQS
emergency

Total impedance of each phase load:
Z＝√(12×12＋16×16)＝√400＝20(Ω)
Line current per phase:
I＝U/Z×1.732＝380/20×1.732≈32.9(A)
Active power:
P＝(U/Z)×(U/Z)×R×3＝(380/20)×(380/20)×12×3＝12996(W)≈13KW
Apparent power:
S＝U×U/Z×3＝380×380/20×3＝21660(Va)＝21.66Kva
Reactive power:
Q＝(U/Z)×(U/Z)×XL×3＝(380/20)×(380/20)×16×3＝17328(Var)≈17.3Kvar

(3) When the voltage of 120V is applied to a load, the current is 5A. How much power does the load consume? (4) "120V, 40W" incandescent lamp has more current
(5) "What is the resistance of a 120V, 1kW consumer?

(3) When 120V voltage is applied to a load, the current is 5A. How much power does the load consume?
——P=UI=600W
(4) What is the current in the "120V, 40W" incandescent lamp?
——From P = IU, I = P / u = 0.333a
(5) "What is the resistance of a 120V, 1kW consumer?
——From P = UI, I = u / R, r = u ^ 2 / P = 14.4 Ω
For reference only!